= Solution
Put $r=v+\sigma/\sqrt2$. Direct differentiation of $\varphi=re^{i\pi/f}$ cancels the mixed radial-angular terms, leaving
$$
|\partial\varphi|^2=\frac12(\partial\sigma)^2+\frac{r^2}{f^2}(\partial\pi)^2.
$$
Use $\lambda=2m^2/v^2$ from part b to expand the potential. The complete local expression is
$$
\boxed{\begin{aligned}
\mathcal L={}&\frac12(\partial\sigma)^2+\frac{v^2}{f^2}(\partial\pi)^2-m^2\sigma^2
-\frac{m^2}{\sqrt2v}\sigma^3-\frac{m^2}{8v^2}\sigma^4\\
&+\frac{\sqrt2v}{f^2}\sigma(\partial\pi)^2
+\frac1{2f^2}\sigma^2(\partial\pi)^2+\frac{m^2v^2}{2}.
\end{aligned}}
$$
The radial <field> is canonically normalized, giving $m_\sigma^2=2m^2$. For arbitrary $f$, the canonically normalized angular <field> is $\pi_c=\sqrt2v\,\pi/f$ and has zero <mass>:
$$
\boxed{m_\sigma=\sqrt2\,m,\qquad m_{\pi_c}=0.}
$$
The interactions comprise radial cubic and quartic self-couplings, a derivative $\sigma\pi\pi$ <coupling> and a derivative $\sigma\sigma\pi\pi$ <coupling>. In canonical variables the latter coefficients are $1/(\sqrt2v)$ and $1/(4v^2)$ respectively. The additive constant shifts the <vacuum> <energy> and does not affect these <masses> or interactions.
The original phase <symmetry> becomes the exact shift $\pi\mapsto\pi+f\alpha$, with $\sigma$ unchanged. Any undifferentiated polynomial $\pi^3$ or $\pi^4$ would violate this shift, so such terms are absent. This argument allows derivative interactions; it forbids an angular potential. The massless angular excitation is the <Goldstone boson> for the broken global $U(1)$.
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